18.4Sum (Map)

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

思路I: 用求3Sum的方法,外加一次for循环,时间复杂度O(n3)

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int size = nums.size();
        if(size < 3) return result;
        
        sort(nums.begin(), nums.end());
        for(int i = 0; i < size-3; i++){
            if(i > 0 && nums[i]==nums[i-1])  continue; //The solution set must not contain duplicate=>no duplicate in the same position
            for(int j = i+1; j < size-2; j++){
                if(j> i+1 && nums[j]==nums[j-1]) continue; //The solution set must not contain duplicate=>no duplicate in the same position
                find(nums, j+1, size-1, target-nums[i]-nums[j], i, j);
            }
        }
        return result;
    }
    void find(vector<int>& nums, int start, int end, int target, int& index1, int& index2){
        int sum;
        while(start<end){
            sum = nums[start]+nums[end];
            if(sum == target){
                item.clear();
                item.push_back(nums[index1]);
                item.push_back(nums[index2]);
                item.push_back(nums[start]);
                item.push_back(nums[end]);
                result.push_back(item);
                do{ //The solution set must not contain duplicate=>no duplicate in the same position
                    start++;
                }while(start!= end && nums[start] == nums[start-1]);
                do{ //The solution set must not contain duplicate=>no duplicate in the same position
                    end--;
                }while(end!=start && nums[end] == nums[end+1]);
            }
            else if(sum>target){
                do{ //The solution set must not contain duplicate=>no duplicate in the same position
                    end--;
                }while(end!=start && nums[end] == nums[end+1]);
            }
            else{//The solution set must not contain duplicate=>no duplicate in the same position
                do{ 
                    start++;
                }while(start!= end && nums[start] == nums[start-1]);
            }
        }
    }
    
private:
    vector<vector<int>> result;
    vector<int> item;
};

思路II：用hash table。O(N^2)把所有pair存入hash表，pair中两个元素的和就是hash值。那么接下来求4sum就变成了在所有的pair value中求 2sum，这个就成了线性算法了。所以整体上这个算法是O(N^2)+O(n) = O(N^2)。

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int size = nums.size();
        int a, b, c, d;
        vector<vector<int>> result;
        unordered_map<int,vector<pair<int,int> > > mp;
        unordered_map<int,int> cnt; //各个数的数量
        if(size < 3) return result;

        sort(nums.begin(), nums.end());
        for(int i = 0; i < size-1; i++){
            if(i > 0 && nums[i]==nums[i-1])  continue;
            for(int j = i+1; j < size; j++){
                if(j> i+1 && nums[j]==nums[j-1]) continue;
                mp[nums[i]+nums[j]].push_back(pair<int,int>{nums[i],nums[j]});
            }
        }
        
        for(int i = 0; i < size; i++){
            cnt[nums[i]]++;
        }
        
        for(unordered_map<int,vector<pair<int,int> > >::iterator it1=mp.begin();it1!=mp.end();it1++){//遍历map
            unordered_map<int,vector<pair<int,int> > >::iterator it2=mp.find(target - it1->first); //查找map
            if(it2==mp.end()) continue;// not found
            if(it1->first > it2->first) continue; //already checked，去重
            
            for(int i = 0; i < it1->second.size(); i++){//访问map元素
                for(int j = 0; j < it2->second.size(); j++){
                    a = it1->second[i].first; //访问pair元素
                    b = it1->second[i].second;
                    c = it2->second[j].first;
                    d = it2->second[j].second;
                    
                    if(max(a,b) > min(c,d)) continue; //四个数两两组合，有6种情况，这里只取两个最小的数在it1的情况，去重
                    cnt[a]--;
                    cnt[b]--;
                    cnt[c]--;
                    cnt[d]--;
                    if(cnt[a]<0||cnt[b]<0||cnt[c]<0||cnt[d]<0){
                        cnt[a]++;
                        cnt[b]++;
                        cnt[c]++;
                        cnt[d]++;
                        continue;
                    } 
                    
                    cnt[a]++;
                    cnt[b]++;
                    cnt[c]++;
                    cnt[d]++;
                    vector<int> tmp = {a,b,c,d};
                    sort(tmp.begin(),tmp.end());
                    result.push_back(tmp);
                }
            }
        }
        return result;
    }
};